The kernel of the natural morphism f : Z -> Z/nZ,   is (n). This morphism is surjective, and has kernel (n). Application of the theorem now gives the obvious fact that Z/(n) is isomorphic to Z/nZ.

If n is a multiple of m, then there is a morphism

Z/nZ -> Z/mZ,      x + (n) -> x + (m).

This morphism is surjective and its kernel is (m), so the theorem implies that that there is an isomorphism

(Z/nZ)/(m) -> Z/(m).

Let x C. Consider the morphism

f : Q[X] -> Q[x],     f(g) = g(x).

Clearly, f is surjective. Let us determine its kernel. Observe that g Q[X] lies in Ker(f) if and only if x is a zero of g. Thus, Ker(f) = 0 if x is not algebraic.

Otherwise, there is a unique polynomial d Q[X] with leading coefficient 1 of minimal degree such that d(x) = 0. By the extended Euclidean algorithm, it readily follows that Ker(f) = (d).

Similarly to the modular arithmetic case, we find, for g a divisor of f, that

(Q[X]/fQ[X])/(g) -> Q[X]/(g).

Consider the morphism

Z[X] -> Z + Zi,     f(X) = f(i).

Clearly, X² + 1 Ker(f).

On the other hand Z[X]/(X² + 1) is readily seen to be isomorphic to Z + Zi. A close analysis of the proof of the theorem gives that Ker(f) must coincide with (X² + 1).